This is one common tactic to attack an problem. You identify the ugliest, hardest piece and attempt to get rid of it. If you can repeat this procedure and get rid of all the nasties, what should/could do next is usually obvious.

This is just the brute force method. It is obviously an overkill in this case, but I think it is worth to learn it. Put:
$$ f(x)=\sum_{j=0}^{+\infty}T(j)\, x^j. $$
The recurrence relation then gives $f(0)=1$ and:
$$\frac{f(x)}{1-x}=\sum_{j=0}^{+\infty}\left(\sum_{k\leq j}T(k)\right)x_j=\sum_{j=0}^{+\infty}(2T(j)-1)\,x^j = 2f(x)-\frac{1}{1-x},$$
hence:
$$f(x)=\frac{1}{1-2x}=\sum_{j=0}^{+\infty}2^j\,x^j,$$
from which:
$$T(j)=2^j$$
as wanted.